Mathematics Test

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Mathematics Test - 37

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Question 1/10
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If 9 A.M.s and 9 H.M.s are inserted between 2 and 3 and A be any A.M. and H be the corresponding H.M.,then H(5−A)=

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A
10

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B
6

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C
-6

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D
0

Solutions

Let A be the $k^{t h}$ A.M., then $H$ will be the $k^{t h} H . M$ ., $Now, A = 2 + k d = 2 + k [\frac{3 - 2}{10}] = \frac{20 + k}{10}$
$\frac{1}{H} = \frac{1}{2} + k d^{'} = \frac{1}{2} + k [\frac{\frac{1}{3} - \frac{1}{2}}{10}] = \frac{30 - k}{60}$
$∴ A + \frac{6}{H} = 5 ⟹ H (5 - A) = 6$
Hence, option 'B' is correct.
Question 2/10
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The digits of a three-digit number form a G.P. If 400 is subtracted from it, then we get another three-digit number whose digits form an arithmetic series. What is the sum of these two numbers?

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1356

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B
1648

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C
1462

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D
1000

Solutions

Let the digits of a three-digit number be $x, y$ and $z$ and the number be $100 z + 10 y + x,$ where $x, y, z$ are in G.P. $∴ y^{2} = x z$
$\Rightarrow x z$ must be a square number If $x z = 4,$ then $y = 2 \Rightarrow 421$ is the number, but $421 - 400 = 21$ which is not a three digit number. $∴ x z = 9,$ i.e., $x = 1, z = 9$
$∴ y = 3,$ so that $x, y, z$ are in $A . P .$ $∴$ The number is $931 .$
The other number will be $531,$ so that 1,3 and 5 are in $A$ . P. Their sum $= 931 + 531 = 1462$
Question 3/10
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The sum of the infinite series $1 + (1 + \frac{1}{2}) (\frac{1}{3}) + (1 + \frac{1}{2} + \frac{1}{2^{2}}) (\frac{1}{3^{2}}) + \dots \dots \infty$

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A
12/5

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B
9/5

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C
8/5

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D
5/3

Solutions

Let, $S_{\infty} = 1 + (1 + \frac{1}{2}) (\frac{1}{3}) + (1 + \frac{1}{2} + \frac{1}{2^{2}}) (\frac{1}{3^{2}}) + \dots \dots \infty$
Now, $t_{r} = (1 + \frac{1}{2} + \frac{1}{2^{2}} + \dots + \frac{1}{2^{r - 1}}) (\frac{1}{3^{r - 1}}) = (\frac{1 - (1 / 2)^{r}}{1 - (1 / 2)}) (\frac{1}{3^{r - 1}})$
-.. Sum of G.P series ]
$\Rightarrow t_{r} = 6 (\frac{1}{3^{r}} - \frac{1}{6^{r}})$
$S_{\infty} = \sum_{r = 1}^{\infty} t_{r} = 6 \sum_{r = 1}^{\infty} (\frac{1}{3^{r}} - \frac{1}{6^{r}})$
$\Rightarrow S_{\infty} = 6 ([\frac{1 / 3}{1 - (1 / 3)} - \frac{1 / 6}{1 - (1 / 6)}]) \dots .$ [Sum of Infinite G.P series $]$
$\Rightarrow S_{\infty} = \frac{9}{5}$
Question 4/10
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The least value of 1 / 2 + 2 / 3 cosec²θ + 3 /8 sec²θ is

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A
13 / 24

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B
61 / 48

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C
61 /25

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D
61 /24

Solutions

$\frac{1}{2} + \frac{2}{3} {cosec}^{2} θ + \frac{3}{8} \sec^{2} θ$
$= \frac{1}{2} + \frac{2}{3} + \frac{2}{3} \cot^{2} θ + \frac{3}{8} + \frac{3}{8} \tan^{2} θ$
$= \frac{37}{24} + \frac{2}{3} \cot^{2} θ + \frac{3}{8} \tan^{2} θ$
Now, $A M \geq G M$ $\frac{\frac{2}{3} \cot^{2} θ + \frac{3}{8} \tan^{2} θ}{2} \geq \sqrt{\frac{2}{3} \cot^{2} θ \times \frac{3}{8} \tan^{2} θ}$
$\frac{2}{3} \cot^{2} θ + \frac{3}{8} \tan^{2} θ \geq 1$
Therefore, $\frac{37}{24} + \frac{2}{3} \cot^{2} θ + \frac{3}{8} \tan^{2} θ \geq \frac{37}{24} + 1 = \frac{61}{24}$
Question 5/10
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If $a_{1}, a_{2}, a_{3}, a_{4}$ and $b$ are real numbers such that $(a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) b^{2} - 2 (a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4}) b + (a_{2}^{2} + a_{3}^{2} + a_{4}^{2}) \leq 0$ then $a_{1}, a_{2}, a_{3}, a_{4}$ are
the terms of a/an

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A
G.P.

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B
H.P.

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C
A.G.P.

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D
A.P.

Solutions

$(a_{1}^{2} + a_{2}^{2} + a_{3}^{3}) b^{2} - 2 (a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4}) b + (a_{2}^{2} + a_{3}^{3} + a_{4}^{2}) \leq 0$
where, $a_{1}, a_{2}, a_{3}, a_{4}, b \in R$
or, $(a_{1}^{2} b^{2} - 2 a_{1} a_{2} b + a_{2}^{2}) + (a_{2}^{2} b^{2} - 2 a_{2} a_{3} b + a_{3}^{2}) + (a_{3}^{2} b^{2} - 2 a_{3} a_{4} b + a_{4}^{2}) \leq 0$
or, ${(a_{1} b - a_{2})}^{2} + {(a_{2} b - a_{3})}^{2} + {(a_{3} b - a_{4})}^{2} \leq 0$
since all individual terms are positive so, they all are equal to zero separately
or, ${(a_{1} b - a_{2})}^{2} = 0$ or ${(a_{2} b - a_{3})}^{2} = 0$ or ${(a_{3} b - a_{4})}^{2} = 0$
or, $(a_{1} b - a_{2}) = 0$ or $(a_{2} b - a_{3}) = 0$ or $(a_{3} b - a_{4}) = 0$
or, $b = \frac{a_{2}}{a_{1}}$ or $b = \frac{a_{3}}{a_{2}}$ or $b = \frac{a_{4}}{a_{3}}$
or, $\frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}} = \frac{a_{4}}{a_{3}} = b$
$So, a_{1}, a_{2}, a_{3}$ and $a_{4}$ are in G.P. Hence, $A$ is the correct option.
Question 6/10
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The equation of the locus of the points equidistant from the points A(−2,3) and B(6,−5) is

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A
x−y=3

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B
x+y=3

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C
2x+y=3

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D
2x−y=3

Solutions

Let an arbitrary point be $P (x, y)$ , Then it is given that $P A = P B$
or $P A^{2} = P B^{2}$
$(x + 2)^{2} + (y - 3)^{2} = (x - 6)^{2} + (y + 5)^{2}$
$(x + 2)^{2} - (x - 6)^{2} = (y + 5)^{2} - (y - 3)^{2}$
$(2 x - 4) (8) = (2 y + 2) (8)$
$2 x - 4 = 2 y + 2$
$x - 2 = y + 1$
$x = y + 3$
or
$x - y = 3$
Question 7/10
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The distance between the points (2,π/2),(5,tan⁻¹4/3) is

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A
2

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B
3

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C
√13

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D
13

Solutions

Converting from polar to rectangular form. $A \equiv (2, \frac{π}{2})$
Here, $r = 2$ and $θ_{1} = \frac{π}{2}$
Hence, $x_{1} = r \cos θ_{1}$ and $y_{1} = r \sin θ_{1}$
$∴ x_{1} = 0$ and $y_{1} = 2$
so, $(2, \frac{π}{2}) \equiv (0, 2)$
$B \equiv (5, \tan^{- 1} \frac{4}{3})$
Here, $r = 5$ and $\tan θ_{2} = \frac{4}{3}$
$∴ \sin θ_{2} = \frac{4}{5}$ and $\cos θ_{2} = \frac{3}{5}$
Hence, $x_{2} = r \cos θ_{2}$ and $y_{2} = r \sin θ_{2}$
$∴ x_{2} = 3$ and $y_{2} = 4$
So, $(5, \tan^{- 1} \frac{4}{3}) \equiv (3, 4)$
By distance formula, $A B = \sqrt{(3 - 0)^{2} + (4 - 2)^{2}}$
$A B = \sqrt{13}$
Question 8/10
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The distance between (a,π/2) and (3a,π/6) is

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A
7a

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B
√7a

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C
2√7a

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D
7√2a

Solutions

Distance between the points in polar form is $d = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} \cos (θ_{1} - θ_{2})}$
$\Rightarrow d = \sqrt{a^{2} + 9 a^{2} - 6 a^{2} \cos \frac{π}{3}}$
$\Rightarrow d = a \sqrt{7}$
Alternate Method:
Converting polar coordinates $(a, \frac{π}{2})$ to cartesian form $x_{1} = a \cos \frac{π}{2}$ and $y_{1} = a \sin \frac{π}{2}$
So, the cartesian form of $(a, \frac{π}{2})$ is $(0, a)$ Converting polar coordinates $(3 a, \frac{π}{6})$ to cartesian form $x_{2} = 3 a \cos \frac{π}{6}$ and $y_{2} = 3 a \sin \frac{π}{6}$
$x_{2} = \frac{3 a \sqrt{3}}{2}, y_{2} = \frac{3 a}{2}$
So, the cartesian form of $(3 a, \frac{π}{6})$ is $(\frac{3 a \sqrt{3}}{2}, \frac{3 a}{2})$ Distance $d = \sqrt{\frac{27 a^{2}}{4} + \frac{a^{2}}{4}}$
$\Rightarrow d = \sqrt{7 a^{2}}$
$\Rightarrow d = a \sqrt{7}$
Question 9/10
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The polar equation of xcos(α)+ysin(α)=p is

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A
r cos(θ−α) =p

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B
r cos(θ+α) =p

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C
r sin(θ−α) =p

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D
r sin(θ+α) =p

Solutions

$x = r \cos θ$ and $y = r \sin θ$ where $r = \sqrt{x^{2} + y^{2}}$
Hence
$x \cos α + y \sin α = p$
$r \cos θ \cos α + r \sin θ \sin α = p$
$r (\cos θ \cos α + \sin θ \sin α) = p$
$r \cos (θ - α) = p$
This is the required polar equation of the given line.
Question 10/10
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The quadratic equation whose roots are the x and y intercepts of the line passing through (1,1) and making a triangle of area A with the axes, may be

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A
x²+ Ax + 2A=0

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B
x² - 2Ax + 2A = 0

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C
x² - Ax + 2A = 0

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D
none of these

Solutions

Let the roots be $α, β$ Then the equation of the line will be $\frac{x}{α} + \frac{y}{β} = 1$ where $α$ is the $x$ intercept and $β$ is the $y$ intercept. Also, the area of the right triangle formed by the line and the coordinate axes is $\frac{α \cdot β}{2} = A$
Or $α \cdot β = 2 A \dots (i)$
since the line passes through ( 1,1 ) hence $\frac{1}{α} + \frac{1}{β} = 1$
$\Rightarrow \frac{α + β}{α \cdot β} = 1$
$\Rightarrow α + β = α \cdot β$
$\Rightarrow α + β = 2 A \dots$ from $i$
Hence, sum of roots is $= α + β = 2 A$
And, Product of roots $α β = 2 A$ Thus the equation with the roots $α$ and $β$ is $(x - α) (x - β) = 0$
$\Rightarrow x^{2} - (α + β) x + α β = 0$
$\Rightarrow x^{2} - 2 A x + 2 A = 0$

Question 1/10

10

6

-6

0

Question 2/10

1356

1648

1462

1000

Question 3/10

12/5

9/5

8/5

5/3

Question 4/10

13 / 24

61 / 48

61 /25

61 /24

Question 5/10

G.P.

H.P.

A.G.P.

A.P.

Question 6/10

x−y=3

x+y=3

2x+y=3

2x−y=3

Question 7/10

2

3

√13

13

Question 8/10

7a

√7a

2√7a

7√2a

Question 9/10

r cos(θ−α) =p

r cos(θ+α) =p

r sin(θ−α) =p

r sin(θ+α) =p

Question 10/10

x2+ Ax + 2A=0

x2 - 2Ax + 2A = 0

x2 - Ax + 2A = 0

none of these

x²+ Ax + 2A=0

x² - 2Ax + 2A = 0

x² - Ax + 2A = 0